Order of (f°h) Group Homework Solution

[RJLiberator]

Homework Statement

Let f,h ∈S_4 be described by:
f(1) = 1
f(2) = 4
f(3) = 3
f(4) = 2

h(1) = 4
h(2) = 3
h(3) = 2
h(4) = 1

Express (f°h) in terms of its behavior on {1,2,3,4} and then find the order of (f°h).

Homework Equations

The Attempt at a Solution

So, first I express (f°h) in terms of its behavior.
(f°h)(1) = 2
(f°h)(2) = 3
(f°h)(3) = 4
(f°h)(4) = 1

Done!
Now, find the order of (f°h):

The order of (f°h) is min{K:g^k-e} or infinity if no such k exists.

I'm having a bit of difficulty in this. To me, when I search wikipedia: https://en.wikipedia.org/wiki/Order_(group_theory)
It says that the order is the cardinality or the number of elements in the set.

(f°h)'s order thus is 4?
But I have a feeling it might be 3 based on in class examples.
My answer is 4.

How did I do?

 

[Orodruin]

You need to separate the order of a group, i.e., the number of elements it contains, from the order of a group element, i.e., the power you need to raise it to to get the identity element.

I would also suggest using cycle notation for group elements of $S_n$, it reduces what you have to write significantly.

 

[RJLiberator]

So the number of elements it contains is 4.
The order of a group element (the power needed to raise it to get the identity element) is 1?
so 4-1 = 3?

 

[Orodruin]

So the number of elements it contains is 4.
The order of a group element (the power needed to raise it to get the identity element) is 1?
so 4-1 = 3?

No, the point is not to mix the concepts. The group is $S_4$ and its order is $4!$, but that was not what the question was about. The question is about the order of the group element fh.

 

[RJLiberator]

Ah. So that is the distinction.
I'm sorry, I made a typo in the question f,h was supposed to be (f°h)

The order of the group element (f°h) is thus 1 as any element in (f°h) has a power of 1.

 

[Orodruin]

Ah. So that is the distinction.
I'm sorry, I made a typo in the question f,h was supposed to be (f°h)

The order of the group element (f°h) is thus 1 as any element in (f°h) has a power of 1.

No. The order of a group element $f$ is the lowest number $k$ such that $f^k = e$, where $e$ is the identity element (which is the only element of order one!).

 

[RJLiberator]

Hm.
OH. The order is 4? Because (f°h)(4) = 1!

That's the e, the identity.

I think the light went off in my head. Correct?

 

[Orodruin]

The order is 4, but the reason is not the one you stated. Try to figure out the order of f and h to start with. Note that also h(4)=1.
Edit: Are you familiar with cycle notation? It will simplify your life significantly.

 

[RJLiberator]

Damn.

The order of h is 4.
The order of f is 1.

f composed of h is thus 4*1 = 4 ?

 

[Orodruin]

Damn.

The order of h is 4.
The order of f is 1.

f composed of h is thus 4*1 = 4 ?

No. What is h^2?

 

[Orodruin]

And again note that the only element with order one is the identity, for no other element $f$ is $f^1=f=e$.

 

[RJLiberator]

So it deals with powers ?

h^2 = I am not even really sure how to express this. Perhaps this is my problem.

h^2 = h(2)^2 = 9 ?

 

[Orodruin]

So it deals with powers ?

h^2 = I am not even really sure how to express this. Perhaps this is my problem.

h^2 = h(2)^2 = 9 ?

No, h^2 is the usual notation for h×h, where × is the group operation.

 

[RJLiberator]

Ok.
h^2 then is

(h°h)(1) = 1
(h°h)(2) = 2
(h°h)(3) = 3
(h°h)(4) = 4

(f°f) is similar.

 

[Orodruin]

Right, so what is the order of f and h? If you apply the same reasoning to f×h, what is its order?

 

[RJLiberator]

I'm looking at this answer: http://math.stackexchange.com/questions/972057/calculating-the-order-of-an-element-in-a-group

And can see they used (1,2,3,4)^4 as an example.

But I don't get how (1,2,3,4)^4 = 1

is it because:
1^4 = 1
2^4 = 16
3^4 = 64
4^4 = 256
And thus all mod 4 = 1 ?

I think that makes sense to me now.

 

[Orodruin]

No, taking a power of the group operation has nothing to do with taking a power of the elements you rearrange. Still, I suggest you first figure out the orders of f and h.

 

[RJLiberator]

Oh, it deals with permutations (from reading the link?)

{1, 2, 3, 4} = the permutations are thus for the power of say, 2 we get {(1,3)(2,4)}
And so the permutations of 4 is the identity back.

 

[Orodruin]

So what are the orders of the elements f, h, and f×h?

 

[RJLiberator]

f = {1, 4, 3, 2}
so the order is 3
due to permutations

h = {4, 3, 2, 1}
so the order is 4

fxh = { 2, 3, 4, 1}
so the order is 4

 

[Orodruin]

f = {1, 4, 3, 2}
so the order is 3
due to permutations

No. What is f^2?
(Cycle notation for f would be f=(24)(1)(4), or the short-hand (24))

h = {4, 3, 2, 1}
so the order is 4

No, what is h^2?
(Cycle notation h = (14)(23))

 

[RJLiberator]

f^2 = (1,3)(4,2)
h^2 = (4,2)(3,1)

This is due to permutations, correct?

 

[Orodruin]

f^2 = (1,3)(4,2)
h^2 = (4,2)(3,1)

This is due to permutations, correct?

No, this is incorrect. What is f(f(1))?

 

[RJLiberator]

Well f(f(1)) is 1.

That much I should know to be true.

 

[Orodruin]

And f(f(2)), f(f(3)), and f(f(4))?

 

[RJLiberator]

f(f(1)) = 1
f(f(2)) = 2
f(f(3)) = 3
f(f(4)) = 4

So f^2 means 2 = the order of f !?
And thus, h^2 means 2 is the order of f.
so 2*2 = 4 for f*h ?

 

[Orodruin]

f(f(1)) = 1
f(f(2)) = 2
f(f(3)) = 3
f(f(4)) = 4

So f^2 means 2 = the order of f !?
And thus, h^2 means 2 is the order of f.
so 2*2 = 4 for f*h ?

No, you are just guessing here. That f(f(x)) = x means that f^2 is the identity and therefore f is of order 2. The order of a product is generally not the product of the orders.

 

[RJLiberator]

My thought was that since f(f(x)) produced the set {1,2,3,4} that this meant it was the order. Precisely what you mean here:

That f(f(x)) = x means that f^2 is the identity and therefore f is of order 2.

So instead of guessing at the product property that doesn't exist. Let's see where we can go with this.
if f = 2 and h=2 in terms of orders

We try to find order of f*h

So, first we find f*h = 2, 3, 4, 1
So we do the process again for order = 2. This results in the set {3, 4, 1, 2}
This is not what we want so we check order = 3. This results in the set {4, 1, 2, 3}
So we check n = 4. This results in {1, 2, 3, 4}

Walouh!

 

[Orodruin]

Right, this is even more evident in cycle notation where f×h = (1234), which is a single cycle containing 4 elements, meaning it will go to the identity when taken 4 times.

 

[RJLiberator]

Thanks for your dedicated help here. A simple operation, but quite confusing the first time seeing it for me. (we just started groups this week).